83 Short Legs, Bulging Eyes

QUIZ2-2 1. A line of flies, known as screwed, is pure-bred for four mutations: lg, which results in the fly having short legs; at, which prevents the formation of antennae; bg, which causes the eyes to bulge out. A screwed male is crossed to a wild type female and all the progeny are phenotypically wild type. A female from this cross is crossed with a screwed male and the following progeny are produced:

282 Wild type 126 Short legs 77 No antennae 16 Short legs, no antennae 24 Bulging eyes 83 Short legs, bulging eyes 134 No antennae, bulging eyes 258 Short legs, no antennae, bulging eyes a) Draw a linkage map of these four genes on the “chromosome” below, including map distances. lg-at = (126+77+83+134)/1000=0.42 lg-bg = (126+16+24+134)/1000=0.30 at-bg= (69+1+8+15+21+6+3+77)/1000=0.20

at-----20-----bg-----30------lg

b) Calculate the interference between regions A & B (a fraction is OK)

COC=[(16+24)/1000]/(0.20*0.30)=0.04/0.06=2/3

I=1-2/3=1/3

c) If interference were 0, what progeny would you expect and in what proportions from the above cross?

at,lg=0.20*0.30/2=0.03 bg=0.03 at,bg=0.30/2-0.03=0.12 lg=0.12 bg,lg=0.20/2-0.03=0.07 at=0.07 at,bg,lg=(1- 0.06-0.24-0.14)/2=0.28 wt=0.28

2. A cross is made between two pure breeding flowers, one with red flowers and long stems, one with blue flowers and short stems. The F1's are all red with long stems. The F1's are selfed and the following phenotypes are observed for 1600 progeny.

red/long 683 red/short 185 blue/long 180 blue/short 552

Set up a Chi squared analysis to determine whether the genes for flower color and stem length are linked or unlinked. Be sure to state the null hypothesis and the degrees of freedom.

Answer: Null hypothesis = unlinked the expected ratio is 9:3:3:1, which for the 1600 would be 900 red/long 300 red/short 300 blue/long and 100 blue/short.

((900-683)^2)/900 +((300-185)^2)/300 + ((300-180)^2)/300 + ((100-552)^2)/100 = Chi squared degrees of freedom = 3

3. Two haploid yeast are mated and allowed to sporulate: his2 trp3 X HIS2 TRP3 There following tetrads are observed:

105 5 46 44 his2 trp3 his2 TRP3 his2 TRP3 HIS2 TRP3 HIS2 TRP3 his2 TRP3 HIS2 trp3 his2 trp3 his2 trp3 HIS2 trp3 his2 trp3 HIS2 trp3 HIS2 TRP3 HIS2 trp3 HIS2 TRP3 his2 TRP3

Determine whether the genes are linked, and if so, calculate the distance between the two genes.

Answer:

PD = 105 NPD = 5 T= 46 + 44

PD>>NPD => linked map distance = (1/2(46+44)+3(5))/200 * 100 = 30 mu